How do you test the improper integral #intx^2 dx# from #(-oo, oo)# and evaluate if possible?

Question

Ezekiel Alexander · Accepted Answer

It is divergent.

We deal with the indefinite integral as normal: So: # int \ x^2 \ dx = 1/3x^3 \ \ (+c) # Then as we are dealing with an infinite integration limit we use the limit definition to get: # int_(-oo)^(oo) \ x^2 \ dx = [ \ 1/3x^3 \ ]_(-oo)^(oo) # # " " = lim_(n rarr oo)[ \ 1/3x^3 \ ]_(-n)^(n) # # " " = lim_(n rarr oo)1/3(n^3-(-n)^3) # # " " = lim_(n rarr oo)1/3(n^3+n^3) # # " " = lim_(n rarr oo)2/3n^3 # Which is clearly divergent (and therefore undefined)

Cameron Alexander · Answer

The integral:

#int_(-oo)^oo x^2dx#

is divergent.

We have for #t > 0#: #int_(-t)^t x^2dx = [x^3/3]_(-t)^t = t^3/3 +t^3/3 = 2/3 t^3# So: #int_(-oo)^oo x^2dx = lim_(t->oo) int_(-t)^t x^2dx = lim_(t->oo) 2/3 t^3 = +oo# so the improper integral is divergent.

Charles Anctil · Answer

undefined To test the improper integral $ \int_{-\infty}^{\infty} x^2 \, dx $, you would split it into two separate integrals:

$$ \int_{-\infty}^{\infty} x^2 \, dx = \lim_{a \to -\infty} \int_{a}^{c} x^2 \, dx + \lim_{b \to \infty} \int_{c}^{b} x^2 \, dx $$

Then, you would evaluate each integral separately and take their limits as $ a $ approaches negative infinity and $ b $ approaches positive infinity. This involves integrating $ x^2 $ with respect to $ x $ and then evaluating the limits of integration.

$$ \int_{a}^{c} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{a}^{c} $$
$$ \int_{c}^{b} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{c}^{b} $$

After integrating, you would take the limits of the resulting expressions as $ a $ approaches negative infinity and $ b $ approaches positive infinity. If these limits exist, you can then evaluate them to find the value of the improper integral. If the limits do not exist, the integral diverges.